Question Video: Using Self-Inductance to Solve for Potential Difference across a Loop of Wire | Nagwa Question Video: Using Self-Inductance to Solve for Potential Difference across a Loop of Wire | Nagwa

Question Video: Using Self-Inductance to Solve for Potential Difference across a Loop of Wire Physics

The current induced in a loop of wire increases at 0.16 A/s. The loop has a self-inductance of 0.55 H. What is the potential difference across the loop?

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Video Transcript

The current induced in a loop of wire increases at 0.16 amperes per second. The loop has a self-inductance of 0.55 henrys. What is the potential difference across the loop?

Let’s say that this is our loop of wire. The wire carries current, and that current changes over time at a rate of 0.16 amperes per second. This change in current is due to a change in the magnetic flux experienced by the loop. The relationship between how much current changes for a given change in magnetic flux is called the self-inductance of the loop. Self-inductance is represented using a capital 𝐿, and the units of self-inductance are henrys. One henry is defined as a weber, the unit of magnetic flux, per ampere, the unit of current. We see then that inductance really does relate magnetic flux to current.

In this example, we want to solve for the potential difference that is induced across the loop. In order for current to be induced and to exist in the loop, there must be some potential difference induced across it. That potential difference, which we will represent in general using this script 𝐸 or letter 𝜀, is equal to the self-inductance of a loop multiplied by the change in current in that loop divided by a change in time. We’ve seen that, in our case, Δ𝐼 divided by Δ𝑡 is 0.16 amperes per second and 𝐿 is 0.55 henrys. So then, the potential difference across our loop of wire equals 0.55 henrys times 0.16 amperes per second, and this product is equal to 0.088 volts. This is the potential difference that is induced across the loop.

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